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Question Number 53474 by maxmathsup by imad last updated on 22/Jan/19
calculate ∫_0 ^1   ((5^(2x+1)  −2^(2x−1) )/(10^x )) dx
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{5}^{\mathrm{2}{x}+\mathrm{1}} \:−\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} }{\mathrm{10}^{{x}} }\:{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
∫_0 ^1 ((5^x ×5^x ×5−((2^x ×2^x )/2))/(5^x ×2^x ))dx  5∫_0 ^1 ((5/2))^x dx−(1/2)∫_0 ^1 ((2/5))^x dx  ∣5×((((5/2))^x )/(ln((5/2))))−(1/2)×((((2/5))^x )/(ln((2/5))))∣_0 ^1   =[{5×((((5/2)))/(ln((5/2))))−(1/2)×((((2/5)))/(ln((2/5))))}−{(5/(ln((5/2))))−(1/2)×(1/(ln((2/5))))}]  =[((((25)/2)−5)/(ln((5/2))))−(1/2)×((2/5)/(ln((2/5))))+(1/2)×(1/(ln((2/5))))]  =((15)/(2ln((5/2))))+(1/(2ln((2/5)))){1−(2/5)}]  =((15)/(2ln((5/2))))+(3/(10ln((2/5))))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{5}^{{x}} ×\mathrm{5}^{{x}} ×\mathrm{5}−\frac{\mathrm{2}^{{x}} ×\mathrm{2}^{{x}} }{\mathrm{2}}}{\mathrm{5}^{{x}} ×\mathrm{2}^{{x}} }{dx} \\ $$$$\mathrm{5}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{{x}} {dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{{x}} {dx} \\ $$$$\mid\mathrm{5}×\frac{\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{{x}} }{{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{{x}} }{{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left[\left\{\mathrm{5}×\frac{\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}{{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left(\frac{\mathrm{2}}{\mathrm{5}}\right)}{{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)}\right\}−\left\{\frac{\mathrm{5}}{{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)}\right\}\right] \\ $$$$=\left[\frac{\frac{\mathrm{25}}{\mathrm{2}}−\mathrm{5}}{{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\frac{\mathrm{2}}{\mathrm{5}}}{{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)}\right] \\ $$$$\left.=\frac{\mathrm{15}}{\mathrm{2}{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}+\frac{\mathrm{1}}{\mathrm{2}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)}\left\{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{5}}\right\}\right] \\ $$$$=\frac{\mathrm{15}}{\mathrm{2}{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}+\frac{\mathrm{3}}{\mathrm{10}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)} \\ $$
Commented by malwaan last updated on 23/Jan/19
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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